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Author: Subject: Electrolysis of aqueous acetate salts
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[*] posted on 27-5-2010 at 12:44
Electrolysis of aqueous acetate salts

What is formed by electrolysis of aqueous acetate salts?

I read on wikipedia:
In the laboratory, ethane may be conveniently prepared by Kolbe electrolysis. In this technique, an aqueous solution of an acetate salt is electrolysed. At the anode, acetate is oxidized to produce carbon dioxide and methyl radicals, and the highly reactive methyl radicals combine to produce ethane.

But what is formed at another electrode?

If I perform electrolysis on sodium acetate, will naoh, ethane and hydrogen be formed?

Maybe this question is easy to answer, but please answer.
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[*] posted on 27-5-2010 at 18:11

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[*] posted on 27-5-2010 at 23:02

I did quite some experimenting on this kind of electrolysis and have written a web page about it. It might be an interesting read for you:

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[*] posted on 28-5-2010 at 11:54

Thanks for the link, woelen. Thought I'd read through all your experiments but clearly I missed at least one.

As to the topic at hand, according to one of the books in the sciencemadness library you can also get methanol at the anode. This would be consistent with a CH3 radical combining with an OH- ion (which woelen avoided by making sure his solutions were acidic).
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[*] posted on 29-5-2010 at 09:15

The production of the corresponding one less carbon alcohol is often referred to as a non-Kolbe or Hofer-Moest reaction.

The Kolbe reaction is favored by high current densities and low temperatures, the non-Kolbe products are increased by addition of indifferent electrolytes such as KHCO3 (0.2 M) and Na2SO4 (1+ M).

Smooth electrodes of platinum, iridium, or rhodium give the best results, graphite electrodes give little to no Kolbe product. Most other common electrode materials, with the exception of vitreous carbon, also give poor yields.

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[*] posted on 22-7-2010 at 23:38

if you use carbon as electrod the methyl radical react with water to form methanol....

(sorry for my english...i'm italian...)
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[*] posted on 23-7-2010 at 00:29

I am absolutely certain that I read in a book that the methyl radical does indeed form.
The reaction stated (if I remember correctly) was
CH2C(=O)-O-O-C(=O)CH3 --> 2CO2 + 2CH3.

Peroxy acetic acid is stable, however; I have worked with it (11% solution).
I do not know whether the 'peroxy acetic anhydride' has the ability or time enough to hydrolyze with water (forming peroxy acetic and acetic acids) before its decomposition to any extent.

Somewhat relevant to this, I have found that a 30% solution of hydrogen peroxide becomes unstable when enough acetic anhydride is added (although there was actually just a little concentrated HCl solution in the mixture also). On slow, gradual periodic addition of acetic anhydride, the mixture first did nothing, then I started seeing minute bubbles rising from an insignificant scratch in the bottom of the glass beaker, then adding just a little more Ac2O, more bubbles appeared. No more Ac2O was added, but after only two minutes the bubbling became more violent and the beaker became hot to the touch; eventually "steam" started coming out and a little bit of solution spilled over, even though the beaker was not initially filled even half way. Methyl radicals may have been transiently produced in this reaction.
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[*] posted on 23-7-2010 at 00:52

Diacetyl peroxide does indeed fragment to give methyl radicals. The process occurs in two steps:

CH3C(O)-O-O-(O)CCH3 => 2CH3C(O)-O* => 2CH3* + 2CO2

What does this have to do with Kolbe electrolysis of acetate though?
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